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Proof step 52 from proof infinite-append-prefixnextnli (previous step)

ii(0) = 0, lf(0) = [], ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), ∀ n. lf(n) = nf to ii(n), ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ (nf from ii(n)) to # nextnli0(lf(n), n) = nextnli0(lf(n), n) ⊦ ∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1)

User applied rule all left

And obtained the following new goals

Proof step 53:
ii(0) = 0, lf(0) = [], ∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), ∀ n. lf(n) = nf to ii(n), ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ (nf from ii(n)) to # nextnli0(lf(n), n) = nextnli0(lf(n), n) ⊦ ∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1)

Rule argument

pos: left 3 terms: n1 abort: F discard: T computed: F

Additional information

Used terms: n1